If secθ + tanθ = m(>1), then the value of sinθ is (0° < θ < 90°)

Answer: Option B

$$\begin{array}{rl}& \sec \theta +\tan \theta =m.......(i)\\ & \text{then, }\sec \theta -\tan \theta =\frac{1}{m}......(ii)\\ & Because{\sec }^2\theta -\text{ta}{\text{n}}^2\theta =1\\ & \text{From equation (i)}-\text{(ii)}\\ & 2\tan \theta =m-\frac{1}{m}\end{array}$$

$$\begin{array}{rl}& \text{tan}\theta =\frac{m^2-1}{2m}\\ & \sin \theta =\frac{m^2-1}{m^2+1}\end{array}$$